Register to reply 
Volleyball Problem involving Projectile Motion 
Share this thread: 
#1
Aug2913, 11:30 AM

P: 25

1. The problem statement, all variables and given/known data
A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.73 m directly above the back line, and the ball's initial velocity makes an angle theta = 40 degrees with respect to the ground. 1)In volleyball, it is often advantageous to serve the ball as hard as possible. If you want the ball to land in the opponent's court, however, there is an upper limit on the initial ball speed for a given contact point. At this maximum speed, the ball just barely makes it over the net and then just barely lands in bounds on the back line of the opponent's court. For the contact point given in the previous problems, what is this maximum initial speed? 2)If you hit the ball at this maximum speed, at what angle should you strike it in order to make sure the ball lands in bounds? 2. Relevant equations Projectile motion equations Y = Yo + Vyo*t 0.5*g*t^2 X = Vx*t And the derivative of those equations. 3. The attempt at a solution I sure I can solve number 2 fine, I just need helping starting number 1. Kind of lost on where to begin and what to start with. Any help would be very much appreciated! 


#2
Aug2913, 11:45 AM

P: 36

well you know your final distance in the x direction (length of court) which is 18m and you know the final height in the y direction which is zero. You can set up two equations, one in the x direction:
v[itex]_{x}[/itex]t=s[itex]_{x}[/itex] and one in the y direction: s[itex]_{yf}[/itex]=[itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex]+v[itex]_{y}[/itex]t+s[itex]_{yi}[/itex] v[itex]_{y}[/itex] can be related to v[itex]_{x}[/itex] using the 40° angle given. With that you have two equations and two unknowns, t and V. Solve for V. 


#3
Aug2913, 11:47 AM

PF Gold
P: 2,308




#4
Aug2913, 12:01 PM

P: 1,596

Volleyball Problem involving Projectile Motion
$$Δd_H = \frac{v_R^2 sin(2θ)}{a}$$ 


#5
Aug2913, 11:05 PM

P: 25

CAF123, I will try that and see where I get. As to Zondrina and abrewmaster, thanks for the help but not sure those methods will work. Even though an angle is given in the problem, that was for previous parts not listed. For this one I need to find the maximum velocity and the angle that would produce it. im just re using the length of court, height of net, and height of contact point.



#6
Aug2913, 11:09 PM

P: 25

I feel like there might be another way at going about this. 


#7
Aug3013, 02:26 AM

PF Gold
P: 2,308




#8
Aug3013, 06:37 AM

Mentor
P: 11,689

With three given points along the trajectory (launch, net top, far baseline), the trajectory is pretty much nailed down; one parabola will pass through those points. So really the idea of "maximum speed" is superfluous to solving for the required speed and angle.
Just write the pair of equations (x and y w.r.t. time) for the two known locations after launch. You can eliminate the time parameters as usual using the xposition equations. You'll be left with two equations in two unknowns (launch angle and launch speed). 


#9
Aug3013, 07:11 AM

P: 223

**When you serve, not accounting for atmospheric resistances right now, the ball will make a trajectory that is symmetric. You serve at 40 deg, it will land at a 40 deg angle. The peak of the ball's height will have to hit right before you reach the position of the net, assuming you serve as stated in the given information, when falling it will hit the ground just before the back line. You can also serve at position0, reach the peak height at 9m horizontal travel distance, then it will touch the back line of the opponents' court.
EDIT: ** As stated, it is NOT TRUE when the starting elevation is not the same as the landing one, the explanation was meant in general as if you were serving from ground level. Now you serve at a height of 1.73m, so primary school volleyball it seems  when you use the same logic as I described, the ball will land over the back line of their court. Imagine the 1.73m being another ground level  the ball travels the same horizontal distance in relation to the angle and the velocity, but in the end it has an extra 1.73m drop time to the actual ground level, which gives it extra distance. Therefore the peak height of the ball will be quite a bit before the net position. I will show you a glimpse of how I would tackle this problem: Let's call the peak height of the ball Hmax. Therefore to reach Hmax we have: Y0 + Vy  gt^2/2 = Hmax Now how would you get the time? When you strike the ball at an angle, you will give it a speed of some sort. That speed is divided into 2 components: the horizontal component and a vertical one. Let's call them Vx and Vy  the V is a hypotenuses of a rightangled triangle so how will you find the sides of the triangle? hmm? :) Which side are you interested in right now? The vertical one, because we need the vertical speed component.
Spoiler
We get V*sin40(deg) as Vy.
Now how about the time? Well, you know that the ball travels a total of 18m horizontally (that is if you seek to hit the backline of their court, which still counts as "in") and in our current conditions, the horizontal speed component will be constant. Therefore Sx = Vx * t. You can substitute Vx as something else, similar like we did to the vertical speed component. t = 18m/V*cos40(deg). This is to just give you an idea where it's going  I won't write out everything and solve it for you, but you should be able to see where it is going by now. Once you get to the 2nd part of the assignment where you have a found the velocity it is already the maximum velocity you can give it at 40 deg. If you need to play around with numbers, you can change your starting angle and find the speed OR show at which angle will you even get across the net and all that stuff. The thing, this method and assignment for that matter, will only work in a 2 dimension world. In reality you also have a choice of direction and that will also affect the maximum allowed speed. Also, bear in mind: I haven't mentioned everything you need to know, yet, maybe you'll ask about them yourself :) 


#10
Aug3013, 07:25 AM

Mentor
P: 11,689




#11
Aug3013, 08:01 AM

P: 223

Yes I pressed the post button accidentally before I got down to the root of the problem, sorry. You gave away a huge hint with your illustration, though, it should help the cause :)
Another point to make is that you will want to reach the least possible angle. You can do it the same like I explained in my earlier post where I started to disect the assignment, but now change the variable to an angle and express distances through velocity. You don't want to hit the ball in the net, but you don't want to fly too high over, for it would mean you are forced to use less power in your serve otherwise you cross their back line. Find the equation you need to express the ball's motion and express it as a function. When your angle changes to the least possible, you will gain in speed, therefore you are more likely to score since the ball is more difficult to stop. A function has 2 extremes, a minimum and a maximum  the angle is your argument, making speed the function. Express it as such and find where your function reaches maximum value. 


#12
Aug3013, 10:37 AM

P: 25

Thanks guys! I solved it now! The help really worked, I just needed to write those equations and visualize the constraints so I could maximize properly



Register to reply 
Related Discussions  
Projectile Motion test problem, involving unk. v0 and y displacement.  Introductory Physics Homework  3  
Projectile motion (the dreaded volleyball problem)  Introductory Physics Homework  17  
Projectile Motion  Volleyball  Introductory Physics Homework  5  
Shotput problem involving projectile motion  Introductory Physics Homework  4  
Volleyball projectile motion  Introductory Physics Homework  3 